The position of a particle moving along a coordinate line is s = sqrt(1 + 4t), with s in meters and t in seconds. Find the particle’s velocity and acceleration at t = 6 sec.

To find the particle's velocity and acceleration at \( t = 6 \) seconds, we need to calculate the first and second derivatives of the position function \( s(t) \) with respect to time \( t \).

Given:
\[
s(t) = \sqrt{1 + 4t}
\]

### 1. Velocity (\( v(t) \)):
The velocity is the first derivative of the position function \( s(t) \) with respect to \( t \):
\[
v(t) = \frac{ds}{dt}
\]

To differentiate \( s(t) = \sqrt{1 + 4t} \), we use the chain rule:
\[
v(t) = \frac{d}{dt} \left(\sqrt{1 + 4t}\right)
\]
\[
v(t) = \frac{1}{2\sqrt{1 + 4t}} \cdot \frac{d}{dt}(1 + 4t)
\]
\[
v(t) = \frac{1}{2\sqrt{1 + 4t}} \cdot 4
\]
\[
v(t) = \frac{4}{2\sqrt{1 + 4t}} = \frac{2}{\sqrt{1 + 4t}}
\]

Now, plug in \( t = 6 \) seconds:
\[
v(6) = \frac{2}{\sqrt{1 + 4(6)}} = \frac{2}{\sqrt{1 + 24}} = \frac{2}{\sqrt{25}} = \frac{2}{5} \text{ meters/second}
\]

### 2. Acceleration (\( a(t) \)):
The acceleration is the first derivative of the velocity function \( v(t) \) with respect to \( t \), which is also the second derivative of the position function \( s(t) \):
\[
a(t) = \frac{dv}{dt}
\]

Differentiate \( v(t) = \frac{2}{\sqrt{1 + 4t}} \):
\[
a(t) = \frac{d}{dt} \left(\frac{2}{\sqrt{1 + 4t}}\right)
\]
\[
a(t) = 2 \cdot \frac{d}{dt} \left((1 + 4t)^{-\frac{1}{2}}\right)
\]
\[
a(t) = 2 \cdot \left(-\frac{1}{2}(1 + 4t)^{-\frac{3}{2}} \cdot \frac{d}{dt}(1 + 4t)\right)
\]
\[
a(t) = 2 \cdot \left(-\frac{1}{2}(1 + 4t)^{-\frac{3}{2}} \cdot 4\right)
\]
\[
a(t) = -\frac{4}{(1 + 4t)^{\frac{3}{2}}}
\]

Now, plug in \( t = 6 \) seconds:
\[
a(6) = -\frac{4}{(1 + 24)^{\frac{3}{2}}} = -\frac{4}{25^{\frac{3}{2}}} = -\frac{4}{125} \text{ meters/second}^2
\]

### Final Answers:
- **Velocity at \( t = 6 \) sec**: \( v(6) = \frac{2}{5} \) meters/second.
- **Acceleration at \( t = 6 \) sec**: \( a(6) = -\frac{4}{125} \) meters/second².

Would you like further details or have any questions?

### Related Questions:
1. How would the velocity and acceleration change if the position function were \( s(t) = \sqrt{1 + 6t} \)?
2. What is the significance of a negative acceleration in this context?
3. How can you find the time when the particle is momentarily at rest?
4. What is the total distance traveled by the particle in the first 6 seconds?
5. How would the position function change if the particle started at a different initial position?

### Tip:
Always remember that velocity is the rate of change of position, and acceleration is the rate of change of velocity. This relationship is crucial in understanding motion.

Learn how to find the velocity and acceleration of a particle moving along a coordinate line at t = 6 seconds using calculus. Detailed step-by-step solution provided.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation